Z Scores and The Standard Normal Curve

Z Scores and The Standard Normal Curve

Properties of the Normal Distribution:

Properties of the Normal Distribution:

Theoretical construction Also called Bell Curve or Gaussian Curve Perfectly symmetrical normal distribution The mean (µ) of a distribution is the midpoint of

the curve The tails of the curve are infinite Mean of the curve = median = mode The “area under the curve” is measured in

standard deviations (σ) from the mean (also called Z).

Total area under the curve is an area of 1.00

Theoretical construction Also called Bell Curve or Gaussian Curve Perfectly symmetrical normal distribution The mean (µ) of a distribution is the midpoint of

the curve The tails of the curve are infinite Mean of the curve = median = mode The “area under the curve” is measured in

standard deviations (σ) from the mean (also called Z).

Total area under the curve is an area of 1.00

Properties (cont.)Properties (cont.) Has a mean µ = 0 and standard deviation σ = 1. General relationships: ±1 σ = about 68.26%

±2 σ = about 95.44%±3 σ = about 99.72%*

*Also, when z=±1 then p=.68, when z=±2, p=.95, and when z=±3, p=.997

Has a mean µ = 0 and standard deviation σ = 1. General relationships: ±1 σ = about 68.26%

±2 σ = about 95.44%±3 σ = about 99.72%*

*Also, when z=±1 then p=.68, when z=±2, p=.95, and when z=±3, p=.997

-5 -4 -3 -2 -1 0 1 2 3 4 5

68.26%

95.44%

99.72%

-5 -4 -3 -2 -1 0 1 2 3 4 5

68.26%

95.44%

99.72%

Z-ScoresZ-Scores Are a way of determining the position of a single score

under the normal curve. Measured in standard deviations relative to the mean of

the curve. The Z-score can be used to determine an area under the

curve which is known as a probability.

Formula:

The formula changes a “raw” score to a standardized score (Z). Table A can then be used to determine the area (i.e. the proportion) beyond z, the area between the mean and z, or the area below z.

Are a way of determining the position of a single score under the normal curve.

Measured in standard deviations relative to the mean of the curve.

The Z-score can be used to determine an area under the curve which is known as a probability.

Formula:

The formula changes a “raw” score to a standardized score (Z). Table A can then be used to determine the area (i.e. the proportion) beyond z, the area between the mean and z, or the area below z.

Using Table A to Find the Area Beyond Z

Using Table A to Find the Area Beyond Z

Table A (back of the book) can be used to find the area beyond the part of the curve cut off by the z-score.

Column C is the probability associated with the area of the curve beyond z (more extreme).

Column B is the proportion of the curve between the z-score and the mean.

The probability of falling within ±z standard deviations of the mean is found by doubling Column B.

Table A (back of the book) can be used to find the area beyond the part of the curve cut off by the z-score.

Column C is the probability associated with the area of the curve beyond z (more extreme).

Column B is the proportion of the curve between the z-score and the mean.

The probability of falling within ±z standard deviations of the mean is found by doubling Column B.

Example: Proportion Below the Mean

Example: Proportion Below the Mean

3. Use Column C

1.

2.

15.87 % of customers

4.

Going the other wayGoing the other way Example: Say a graduate school

only admits students with GRE math scores in the top 10% (mean = 500, standard deviation = 100). What score does one need to be admitted? Like last time but in reverse: from the

proportion (10%, or .1000), look up Z in the table (column C): Z = 1.28. Solve for x! Z=1.28=(x-500)/100. So, x =

628.

Example: Say a graduate school only admits students with GRE math scores in the top 10% (mean = 500, standard deviation = 100). What score does one need to be admitted? Like last time but in reverse: from the

proportion (10%, or .1000), look up Z in the table (column C): Z = 1.28. Solve for x! Z=1.28=(x-500)/100. So, x =

628.

Using Table A (cont.)Using Table A (cont.) Example: Say you score 130 on an IQ test with mean =

100 and standard deviation = 15 in the general population. What percentage of the population scores higher than you? Z = (130-100)/15 = +2.00. According to the table,

the area beyond (Z = +2.00) is .0228 (column C), and thus the proportion of the population with scores exceeding this score, would be .0228. So, the answer is 2.28%.

Example: Say you score 130 on an IQ test with mean = 100 and standard deviation = 15 in the general population. What percentage of the population scores higher than you? Z = (130-100)/15 = +2.00. According to the table,

the area beyond (Z = +2.00) is .0228 (column C), and thus the proportion of the population with scores exceeding this score, would be .0228. So, the answer is 2.28%.

Using Table A (cont.)Using Table A (cont.) Example: Say your friend scores 115 on an IQ test with

mean = 100 and standard deviation = 15 in the general population. What percentage of the population scores below him?

This is answered by Column B: if Z = (115-100)/15 = 1.00, then the proportion scoring between the mean of 100 and the score of 115 is .3413. We must add .5 to this to account for the half of the population below the mean, arriving at .8413 or 84.13%

Example: Say your friend scores 115 on an IQ test with mean = 100 and standard deviation = 15 in the general population. What percentage of the population scores below him?

This is answered by Column B: if Z = (115-100)/15 = 1.00, then the proportion scoring between the mean of 100 and the score of 115 is .3413. We must add .5 to this to account for the half of the population below the mean, arriving at .8413 or 84.13%

Using Table A (cont.)Using Table A (cont.) Example: Say you score 130 on an IQ test with mean =

100 and standard deviation = 15 in the general population.Say your friend scores 115.

What proportion of the population is smarter than him but not as smart as you? Be careful here: you can’t add/subtract Z scores!

Instead, find the difference in proportions: Your score of 130 (Z=2.00) exceeds all but .0228 of the population; his (115, Z = 1.00), exceeds all but .1587 (use either Column B or C, but use the same for both values). So: .1587-.0228 = .1359, or 13.59% of the population.

Example: Say you score 130 on an IQ test with mean = 100 and standard deviation = 15 in the general population.Say your friend scores 115.

What proportion of the population is smarter than him but not as smart as you? Be careful here: you can’t add/subtract Z scores!

Instead, find the difference in proportions: Your score of 130 (Z=2.00) exceeds all but .0228 of the population; his (115, Z = 1.00), exceeds all but .1587 (use either Column B or C, but use the same for both values). So: .1587-.0228 = .1359, or 13.59% of the population.

Research: Rejection Region

Research: Rejection Region

α=.05 zcritical=+1.96 and zcritical=-1.96

α=.05 zcritical=+1.96 and zcritical=-1.96

Homework ch. 5Homework ch. 5Say the # of candy bars eaten in a month by each

student at this school is normally distributed, with a mean of 15 and a standard deviation of 3.

1. What percentage of students ate: a. more than 21?b. more than 12?c. between 10 and 20?d. either less than 10 or more than 20?

2. How many candy bars would one have to eat to be in:a. The top 5%?b. The bottom 10%?c. The middle 50%?d. At least 2 standard deviations away from the mean?

Say the # of candy bars eaten in a month by each student at this school is normally distributed, with a mean of 15 and a standard deviation of 3.

1. What percentage of students ate: a. more than 21?b. more than 12?c. between 10 and 20?d. either less than 10 or more than 20?

2. How many candy bars would one have to eat to be in:a. The top 5%?b. The bottom 10%?c. The middle 50%?d. At least 2 standard deviations away from the mean?